Sequentially Compact Metric Space is Compact

Theorem

A sequentially compact metric space is compact.

Proof

Proof using Lindelöf Property

Let $(X,d)$ be a sequentially compact metric space. Our proof is based on the result that a sequentially compact metric space is Lindelöf, that is, every open cover of $X$ has a countable subcover.

Take any open cover $C$ of $X$, and extract from it a countable subcover $\{U_1, U_2, \ldots \}$.

Reasoning by contradiction, assume that there is no finite subcover of $C$. Then, for any natural $n \geq 1$, the family $\{U_1,\ldots,U_n\}$ does not cover $X$, so we can choose a point $x_n \in X$ such that

$x_n \notin U_1 \cup \cdots \cup U_n.$

In this way we have constructed an infinite sequence $\{x_n\}_{n \geq 1}$ of points of $X$.

As we are assuming $X$ is sequentially compact, this sequence has a subsequence which converges to some $x \in X$. But there is some $U_m$ such that $x \in U_m$ (as the $U_i, i \geq 1$, are a cover), and hence by one of the characterizations of convergence, there is an infinite of number of terms in the sequence $\{x_i\}$ which are contained in $U_m$.

This is a contradiction, as in the way we constructed our sequence, each $U_n$ can only contain a finite number of the terms ($U_n$ can contain only points $x_i$ with $i < n$).

$\blacksquare$

Proof using Lebesgue Number

Let $M = \left({X, d}\right)$ be a sequentially compact metric space.

Let $\mathcal{U}$ be any open cover of $M$.

By Lebesgue's Number Lemma, there exists a Lebesgue number for $\mathcal{U}$.

By Sequentially Compact Metric Space has Finite Net, there exists $\left\{{x_1, x_2, \ldots, x_n}\right\}$ which is a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.

Then $N_{\epsilon} \left({x_i}\right)$ is contained in some $U_i \in \mathcal{U}$ by definition of Lebesgue number.

Since $\displaystyle M \subseteq \bigcup_{i=1}^n N_{\epsilon} \left({x_i}\right) \subseteq \bigcup_{i=1}^n U_i$, we have a finite subcover $\left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal{U}$ for $M$.

Hence the result.

$\blacksquare$